➡️How to Design Hybrid Solar PV Systems
Hybrid Solar PV Systems:- A hybrid solar PV system is essentially a system that contain more than one power supply other than the PV, to meet the electrical power demand of the loads. Configuration of hybrid solar PV system is shown in below figure.
The system need to design on the basis of following requirements.
A.
Load – 2 No’s of
5 HP Lathe machine
B.
Duration of
operation -14 Hrs. 6 AM -8 PM
C.
Location -
Hyderabad
D.
Hybrid system
E.
Electricity grid
supply not available – 7 AM- 10 AM and 6-8 PM (total 5 Hrs.)
F.
Redundancy of
system – 1 day
G.
Please use
following specs for project
a.
Modules -320 W,
Vikram module
b.
Statcon inverter
/SMA – Hybrid inverter
c.
Exide or eastmen
battery- 200 Ah
d. Raised Structure -12 feet, C type (SS) channel, 2.5 MM thick
☛ Click Here For View Sample of Solar Module Mounting Structure
Design By :- AutoCAD
1.
Solar
Resources Data of Hyderabad City
A. Average Direct Normal Ir-radiance
I.
Hyderabad,
Telangana, India
II.
Latitude
: 17.35
III.
Longitude
: 78.45
IV.
Annual
Average : 5.27 kWh/m2/day
B. Average Global Horizontal Ir-radiance
I.
Hyderabad,
Telangana, India
II.
Annual Average : 5.77 kWh/m2/day
III.
Solar Azimuth Angle 450
Source of above data NREL site
1.
Load
Assessments
We have 2 No’s of 5 HP lathe machine for water
pumping
1
HP = 746 Watts
746 W * 2 = 1,492 W 1.4 Kw
This is the wattage of lathe machine
Hence total loads required for 1 hrs. 1.4 Kwh ≅ 1.5 Unit
Total duration of operation = 19
9 hrs. Power supply through PV + 5 hrs. Power supply
through Battery + Redundancy of system 5 hrs. next day =
19 hrs.
19
* 1.5 = 28.5
Our 1 kw PV system generate per day
4.5 unit electricity
28.5/4.5 = 6.4 approx. we need 7 Kw System.
We are placed
our SPV module on site of angle of place latitude with south orientation.
2.
System
Design & Calculation
As per inverter availability we are connect Statcon hybrid
inverter with rated capacity 7 Kw.
According to question we are using Waaree Solar module of 350 Wp.
System Calculation :
I.
No. of PV module
We are require 20 Module ☛ Click Here For Module Datasheet
II.
No. of panel in
a string
As per inverter
datasheet
MPPT voltage range – 132 to 266
Maximum input DC
voltage range – 106 V
Maximum voltage range
of inverter ÷ Module Voc
266 ÷ 46.40 = 5.73 Approx 6
Hence
6 Module connected in series * 37.50 Vmp of Solar PV Module =225 V this data is lies our MPPT voltage range.
Thus
We are connect 6 module in series.
III.
String Power
Module wattage * No. of
PV module connected in series
350 * 6 = 2,100
Watts
Total wattage of plant /
String power
7,000 ÷ 2,100 = 3.33 ≅ 3 Strings
3.33* 2100 = 6,993 Watts or 6.9 Kw Approx.
≅ 7 Kw
IV.
No. of module in
a sting
Total No. of PV module
/ No. of PV module connected in series
20 ÷ 6 =3.33
No. of strings * 13
module per string * wattage of module should be equal to our plant capacity
3.33 * 6 * 350 =
V.
Conclusion
1.
No. of PV module
-20
2.
No. of Strings -3
3.
No. of module
connected series -6
4.
Total Capacity – 6.9≅ 7 Kw
3.
Battery
Specification
Battery specifications are given below
a.
Battery model –
6LMS200L
b.
Type of battery
– lead acid battery
c.
Class rating C10
d.
12 V
e. 200Ah
4.
Day
Of Autonomy And Battery Bank Size
Day of autonomy
(No. of day system should supply load without sunlight & Grid )
5 hrs. per day + 5 hrs. autonomy day = 10 hrs.
1.5 * 5 = 7.5 ≅8 Unit for today & 1.5 * 5 = 7.5 ≅8 Unit for next day
Total = 16 unit (power need to supply through battery )
Now
Power of one
battery
V * I = P
12* 200 = 2400 Watts or 2.4 unit (Power of
one battery)
16 ÷ 2.4 = 6.66 ≅ 7 battery Require for whole system
But
Battery Has Certain Parameter Like DoD (Depth of Discharge), Here we are consider 50 % of Battery DoD
Again from this Formula
I = P
Here
I = 16000 ÷ 12
= 1,333 ≅ 1350 Ah
Thus = 1350 ÷ 0.5 = 2700 Ah
Now, This Ah Capacity Divide By One Battery Ah, Which is mentioned above 200 Ah
Hence = 2700 ÷ 200 = 13.5 ≅ 14 Battery
14 battery Require for whole system
In this context i also want to explain that A lot of different types of losses are occur in the battery such as Heating Losses,Chemical conversion losses,Wiring Losses etc. I am not calculating all these losses here because I have taken an estimated large number everywhere.
3. Advantage & Disadvantage of This System:-
I.
Advantage
v Continuous power supply
v Utilize the renewable sources in best way
v
Low
maintenance cost
v
High
efficiency
v Load management
II.
Disadvantage
v Complicated controlling process
v High installation cost
v
Less battery life
v Such type system is not feasible economically
v
The number of
instruments connectable is limited
v
Such type system
will have more conversion losses like PV DC current in AC current and two times
losses will be founded twice time in our battery. One in the storing time and
other in energy releasing time.
v Payback
period of such type system is too long
5. Conclusions & Recommendations
v Such
type system is not economical for more than 100 kw capacity because payback
period of such type plant is too long.
v Need to reduce conversion losses.
v Such
type heavy load we can not operate long time through the battery.
A battery has life cycle up to 5 years. Before it our payback period will not complete and we may have to need to replace battery again. In this way cost of our system can be even more expensive
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